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2 years ago

Hello!! I need help please! Please explain how I can identify a percipitate in a reaction! I need help with chemistry ha

1 answer:

5 0

Generally speaking a precipitate is something that will form as a result of the reaction. It generally is a solid material.

Sodium and nitrates are soluble so Calcium Phosphate would precipitate out....

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Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.

goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

$P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}$

In the first solution:

$X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625$

$X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375$

$340torr = 0.625P^0_{A}+0.375P^0_{B}$ (1)

For the second equation:

$X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700$

$X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300$

$370torr = 0.700P^0_{A}+0.300P^0_{B}$ (2)

Replacing (2) in (1):

$340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})$

$340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}$

-122.5torr = -0.250P°A

$P^0_{A} = 490 torr$

Vapour pressure of cyclohexane at 50°C is 490torr

And for benzene:

$370torr = 0.700*490torr+0.300P^0_{B}$

$P^0_{B}=90torr$

Vapour pressure of benzene at 50°C is 90torr

3 0

2 years ago

For the reaction 2 NO(g) + 2 H2(g) => N2(g) + 2 H2O(g) the following data were collected. What is the rate law for this react

harkovskaia [24]

Answer: The rate law expression for the given reaction is written below.

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(g)$

Rate law expression for the reaction:

$\text{Rate}=k[NO]^2[H_2]^2$

Hence, the rate law expression for the given reaction is written above.

5 0

3 years ago

Help me plz, I need help on this.

kirza4 [7]

Omg i lost everything ugh
To do it again

1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol

2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol

3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol

4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol

5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g

6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g

7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g

8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g

I cant believe i had to do this all over

4 0

3 years ago

Ammonia can be produced via the chemicalreaction

otez555 [7]

Answer:

3)The reaction is not at equilibrium and willproceed to the right.

Explanation:

The reaction quotient of an equilibrium reaction measures relative amounts of the products and the reactants present during the course of the reaction at particular point in the time.

It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.

Q < Kc , reaction will proceed in forward direction.

Q > Kc , reaction will proceed in backward direction.

Q = Kc , reaction at equilibrium.

Given that:

Q = $3.56\times 10^{-4}$

K = $6.02\times 10^{-2}$

Since, Q < K , reaction is not at equilibrium and will proceed to right, in forward direction.

3 0

3 years ago

Thea is doing a chemistry experiment. The instructions say she needs to use pure water. Hamza offer her a bottle labelled “100%

Diano4ka-milaya [45]

How is this going to help us in life wat the fawk

5 0

2 years ago